Question: $\begin{aligned} &g(x)=\sqrt[3]{2x-5}+1 \\\\ &h(y)=2-y \end{aligned}$ $h\left(g\left(-\dfrac{3}{2}\right)\right)=$
Solution: When evaluating composite functions, we work our way inside out. To evaluate $h\left(g\left(-\dfrac{3}{2}\right)\right)$, let's first evaluate $g\left(-\dfrac{3}{2}\right)$. Then we'll plug that result into $h$ to find our answer. Let's evaluate $g\left({-\dfrac{3}{2}}\right)$. $\begin{aligned}g(x)&=\sqrt[3]{2x-5}+1\\\\ g\left({-\dfrac{3}{2}}\right)&=\sqrt[3]{2\left({-\dfrac{3}{2}}\right)-5}+1~~~~~~~~~~\text{Plug in }x={-\dfrac{3}{2}}\\\\ &=\sqrt[3]{-8}+1\\\\ &=-2+1\\\\ &={-1}\end{aligned}$ We now know that $h\left(g\left({-\dfrac{3}{2}}\right)\right)$ is the same as $h({-1})$ because $g\left({-\dfrac{3}{2}}\right) = {-1}$. Let's evaluate $h({-1})$. $\begin{aligned}h(y)&=2-y\\\\ h({{-1}})&=2-({-1})~~~~~~~~~~\text{Plug in }y={-1}\\\\ &=2+1\\\\ &=3\end{aligned}$ The answer: $h\left(g\left(-\dfrac{3}{2}\right)\right)=3$